Six cups and saucers come in pairs: there are two cups and saucers that are red, white, and blue. If the cups are placed randomly onto the saucers (one each), find the probability that no cup is upon a saucer of the same color.

Well first of all, I think that the sample space has a size of 5! since there are 6! ways to arrange the cups, but since there are two of each color, we need to divide 6! by 8 to account for these indistinguishable arrangements.

But I am having trouble counting how many ways to place a cup on the wrong saucer. Is there a clever way to do this or should you just do this by enumerating different ways to put cups on wrong saucers?

  • $\begingroup$ I think you should divide by $2 \times 2 \times 2 = 8$, instead of $6$. Also, are the same color saucers equivalent, just like the way you've treated the cups? So if I have an arrangement where there's a white cup and a blue cup on the two red saucers, does swapping them give an equivalent arrangement? $\endgroup$ – Platehead Oct 7 '14 at 3:29
  • $\begingroup$ Sorry- I did mean to divide by 8. Thanks. I suppose swapping them would give an equivalent arrangement. $\endgroup$ – DHH Oct 7 '14 at 3:42
  • $\begingroup$ Then we should divide again to account for swapping the saucers too? (Just clearing up how the question is set up.) $\endgroup$ – Platehead Oct 7 '14 at 3:44
  • $\begingroup$ see math.stackexchange.com/questions/2955019/… for two alternative solutions based on the inclusion-exclusion principle. $\endgroup$ – utobi Oct 15 '18 at 7:32

Suppose we line up the saucers in order as R R W W B B.

Then we have $\binom{6}{2}\binom{4}{2}=\frac{6!}{2!2!2!}=90$ ways to arrange the cups in order on the saucers.

To count the number of ways no cup lands on a saucer of the same color, we can look at 3 cases:

1) If the red cups are on the white saucers, there is only one possible arrangement (since the blue cups must be on the red saucers).

2) If the red cups are on the blue saucers, there is only one possible arrangement (since the white cups must be on the red saucers).

3) If one red cup is on a white saucer and one red cup is on a blue saucer, then there are 2 choices for the white saucer and 2 choices for the blue saucer. The remaining white saucer must have a blue cup, and the remaining blue saucer must have a white cup; so there 2 ways to arrange the blue cup and white cup that are left on the red saucers.

This gives a total of $1+1+2\times2\times2=10$ arrangements with no cup on a saucer of the same color,

so we get a probability of $\frac{10}{90}=\frac{1}{9}$.


Imagine six numbered cups being placed on six numbered saucers. Items (cups or saucers) numbered $1$ or $2$ are red, those numbered $3$ or $4$ are white, and the others are blue. Place all the saucers randomly in a row, then place all the cups randomly on top. There are $6!$ ways to place the saucers, $6!$ ways to place the cups, so altogether $6! \cdot 6!$ arrangements.

But each such arrangement is just one of $6!$ different arrangements that all pair up the cups and saucers the same way. So let's put all the items of one type in predetermined places and only randomly distribute the items of the other type to those places; then we have just $6!$ arrangements to be concerned with.

Now we observe that for every arrangement with cup number $1$ on saucer $x$ and cup $2$ on saucer $y,$ there is a second arrangement with cup $2$ on saucer $x$ and $1$ on $y$ and all other cups and saucers paired the same as in the first arrangement; and both these arrangements have identical patterns of matched and mismatched colors. So we can consider just half as many arrangements if we ignore the different numbers on the two white cups. Having done this, we find we can divide by two again by ignoring the numbers on the red cups, and again by ignoring the numbers on the blue cups. So we end up with just $6!/2^3 = 90$ distinct arrangements.

But having done this, we can't now apply the argument to the saucers too. If saucers $1$ and $2$ both have blue cups on them, swapping these saucers doesn't change the arrangement at all--it's still "blue on $1,$ blue on $2$" (because we've already decided that the difference between the blue cups doesn't matter). On the other hand, if there is a red cup on saucer $1$ and a blue cup on saucer $2,$ then swapping the saucers results in "blue on $1,$ red on $2,$" which we were counting as a distinct arrangement in our previous list of $90$ arrangements.

We could have applied the swapping argument to the saucers first, but then we couldn't have applied it to the cups.

(Also observe that if you do try to divide by $2$ again for each pair of same-colored saucers, you end up with $90/2^3 = \frac{15}{2}$ distinct arrangements, which doesn't make sense. But that's not the reason why we must not do that.)

To count the number of arrangements with no cup on a saucer of the same color, I suggest considering two cases separately:

Case 1. Both white cups are on the same color of saucer.

Case 2. Each white cup is on a different color of saucer.

Case 1 is very easy to count because once you decide where the white cups are, the other four cups have only one possible arrangement. But in Case 2 you have more ways to arrange the white cups to begin with (remembering that we have to count arrangements with a white cup on saucer $3$ and a different color on $4$ separately from arrangements with white on $4$ and a different color on $3$); and having done so, you will find that there are multiple distinct ways to arrange the remaining cups (remembering that we have to count "red on $1,$ blue on $2$" as a different arrangement than "blue on $1,$ red on $2$" despite the fact that saucers $1$ and $2$ are both white). In fact it turns out you can pick just one of the possible arrangements of cups in case 2 and from it you can generate all the other case-2 arrangements by swapping different-colored cups on saucers of the same color.

Alternatively, I think you could work the problem in the other direction, by showing that you can partition all the possible arrangements into $k$ sets of $n$ arrangements each, such that exactly one of the arrangements in each subset has no cup on a saucer of the same color, and therefore the probability is $\frac 1n;$ but you have to be careful about how you define the partitions in order to make sure that you neither leave an arrangement out of all the subsets of your "partition" nor include the same arrangement in two different subsets.


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